Precalculus - notes for 4.7 Inverse Trigonometric Functions

4.7 Inverse Trigonometric Functions

I. The inverse sine function:
y = arc sin x if and only if sin y = x

We will use a capital A for Arcsin to show when the range is restricted so therefore:

y = Arc sin x
Domain: [-1, 1]
Range: restricted to [-π/2, π/2 ]

y = sin x some points are {(- π/2, -1), (0,0), (π/2, 1)}

so for the inverse, switch the x-values and y-values to get

y = Arc sin x some points are {(-1, - π/2), (0, 0), (1, π/2)}

Sketch the graph

Evaluating the inverse sine function:

Example 1: Arc sin (-1/2) = x
x = {-30º, 330º, 210º, - π/6, 11π/6, 7 π/6}

Example 2: Arc sin √(2)/2 = x
x = {45º, 135º, π/4, 3 π/4}

Example 3: Arc sin -√(3)/2 = x
x = {-60º, -120º, 300º, 240º, -π/3, -2π/3, 5π/3, 4π/3}
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II. The inverse cosine function:

y = arc cos x if and only if cos y = x

Again, to show a restricted range:

y = Arc cos x
Domain: [-1, 1]
Range: [0, π]
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II. The inverse tangent function:

y = Arc tan x if and only if tan y = x

Domain: all the real numbers
Range: (-π/2, π/2)
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Recall f ( f^-1 (x)) = x and f^-1 (f (x)) = x

Therefore: sin (Arc sin x) = x
Arc sin (sin x) = x
cos (Arc cos x) = x
Arc cos (cos x) = x
tan (Arc tan x) = x
Arc tan (tan x) = x

This only applies with sine in the interval [-π/2, π/2]
This only applies with cosine in the interval [0, π]
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Example 1: sin (Arc sin 0.7) = 0.7

Example 2: tan (Arc tan 35) = 35

Example 3: cos [ Arc cos (-0.3)] = -0.3

Example 4: Arc sin (sin 3 π)
Recall 3 π is not in the interval [0, π] so sin (3π) = sin (π) = 0 so
Arc sin (0) in the interval [-π/2, π/2] is equal to 0.

Example 5: sin (Arc tan 4/3) =
Using Pythagorean Theorem and since tangent is opposite divided by adjacent. Therefore 4 is the opposite and 3 is the adjacent to the angle. Using Pythagorean Theorem, that means that the hypotenuse is 5. So the sine would equal 4/5.

Example 6: tan (arc cos x/5) = θ
This means that x is the adjacent and 5 is the hypotenuse.

a² + b² = c²
x² + b² = 5²
b² = 25 - x²
b = the square root of (25 - x²)

so the tan θ = (the square root of (25 - x²))/5

Example 7: cot (arc tan 1/x)
This means that the tangent is equal to 1/x ,
tangent/1 = 1/x
taking the reciprocal of both sides gives us
1/tangent = x/1
1/tangent is equal to cotangent so

so the cotangent is equal to x.

Example 8: Arc cos 3/ (the square root of (x² - 2x + 10)) = Arc sin ___

Using Pythagorean Theorem:
3² + b² = (the square root of (x² - 2x + 10))²

9 + b² = x² - 2x + 10
b² = x² - 2x + 1
b² = (x - 1)(x - 1)
b² = (x - 1)²
b = x - 1

Therefore sin θ = (x - 1)/(the square root of (x² - 2x + 10))

so the Arc cos 3/(the square root of (x² - 2x + 10)) = Arc sin ((x - 1)/(the square root of (x² - 2x + 10))

Prove the identity:

Arc sin x + Arc cos x = π/2

Let θ = Arc sin x and β = Arc cos x

By substitution: θ + β = π/2

Now since: Arc sin x = θ and β = Arc cos x
This means sin θ = x and cos β = x.

Since x = x, then sin θ = cos β
For this to happen, the triangle has to be a right triangle.

Therefore θ + β = π/2

Since this is the same answer as our substitution above, we now know

Arc sin x + Arc cos x = π/2

4.7 homework #34; pg. 351; #1 - 15 odd, 19 - 23 odd, 27 - 53 odd, 57, 59, 71, 87, 88
pg. 341; #19, 45, 52, 64, 91
Quiz on Sections 4.5 - 4.6 next class